The following results were obtained during kinetic studies of the reaction; 2A + B - Products. Expe

Q: Why do we ignore B when calculating the order of A?

Because we found that the reaction is zero order with respect to B. This means B's concentration has absolutely no effect on the rate. It's mathematically equivalent to multiplying by 1.

Q: What if the rate didn't exactly double when A doubled?

If the rate quadrupled, it would be second order ($2^2 = 4$). If it increased by a factor of 8, it would be third order ($2^3 = 8$). You always solve the equation $( \text{Factor of Concentration Change} )^n = ( \text{Factor of Rate Change} )$.

Q: Can I use Experiment II and III to find the order of A?

Yes, but it's slightly more math. Since B is zero order, its change from 0.25 to 0.30 doesn't matter. You'd still see A doubling and the rate doubling, leading to the same conclusion.

Q: Why is the unit of the rate constant $\text{min}^{-1}$?

The overall order of the reaction is 1. The rate has units of $\text{M min}^{-1}$ and $[A]$ has units of $\text{M}$. Dividing rate by $[A]$ leaves $\text{min}^{-1}$.

Q: Does the half-life depend on the initial concentration here?

No! For a first-order reaction, the half-life is a constant ($t_{1/2} = \frac{0.693}{k}$). It takes the same amount of time to go from 0.10 M to 0.05 M as it does to go from 0.05 M to 0.025 M.

Question

The following results were obtained during kinetic studies of the reaction;
$2A + B \longrightarrow 	ext{Products}$

\begin{array}{cccc} \hline 	ext{Experiment} & 	ext{[A] (mol L}^{-1}	ext{)} & 	ext{[B] (mol L}^{-1}	ext{)} & 	ext{Initial rate (mol L}^{-1} 	ext{min}^{-1}	ext{)} \ \hline 	ext{I.} & 0.10 & 0.20 & 6.93 	imes 10^{-3} \ 	ext{II.} & 0.10 & 0.25 & 6.93 	imes 10^{-3} \ 	ext{III.} & 0.20 & 0.30 & 1.386 	imes 10^{-2} \ \hline \end{array}

The time (in minutes) required to consume half of $A$ is

Options:

(A) 5

(B) 10

(C) 100

(D) 1

Sigma Percentile · Accepted Answer

### Decoding the Kinetic Data

Chemical kinetics is like being a detective. You are given a set of clues—in this case, a table of experimental data—and your job is to uncover the hidden rules governing the reaction. The fundamental rule we are looking for is the **Rate Law**, which generally looks like this:

$r = k[A]^a[B]^b$

Here, $r$ is the initial rate of the reaction, $k$ is the rate constant, and the exponents $a$ and $b$ represent the order of the reaction with respect to reactants $A$ and $B$. Our primary mission is to find these exponents. To do this, we use the **Method of Initial Rates**, which involves comparing different experiments to isolate the effect of one reactant at a time.

### The Power of Isolation

Let's start by finding the order with respect to $B$. We need to find two experiments where the concentration of $A$ is kept constant, but the concentration of $B$ changes. Looking at the table, Experiments I and II fit this perfectly.

In Experiment I, $[A] = 0.10 	ext{ M}$ and $[B] = 0.20 	ext{ M}$.
In Experiment II, $[A] = 0.10 	ext{ M}$ and $[B] = 0.25 	ext{ M}$.

Notice that while $[B]$ increased by a factor of $1.25$, the initial rate remained completely unchanged at $6.93 	imes 10^{-3} 	ext{ mol L}^{-1} 	ext{min}^{-1}$.

Mathematically, we can set up a ratio:

$\frac{r_2}{r_1} = \left(\frac{[A]_2}{[A]_1}ight)^a \left(\frac{[B]_2}{[B]_1}ight)^b$

$1 = (1)^a \left(\frac{0.25}{0.20}ight)^b$

$1 = (1.25)^b$

The only way a number raised to a power can equal $1$ is if the power itself is zero. Therefore, **$b = 0$**. This tells us a profound physical truth: the reaction is zero-order with respect to $B$. The concentration of $B$ has absolutely no impact on how fast the reaction proceeds.

### Unveiling the Order of A

Now that we know $B$ is a silent spectator in the rate-determining step, we can find the order with respect to $A$. Let's compare Experiments I and III.

In Experiment I, $[A] = 0.10 	ext{ M}$.
In Experiment III, $[A] = 0.20 	ext{ M}$.

The concentration of $A$ has exactly doubled. What happened to the rate? It went from $6.93 	imes 10^{-3}$ to $1.386 	imes 10^{-2}$. If you look closely, $1.386 	imes 10^{-2}$ is exactly twice $6.93 	imes 10^{-3}$.

Setting up our ratio again:

$\frac{r_3}{r_1} = \left(\frac{[A]_3}{[A]_1}ight)^a$

$\frac{1.386 	imes 10^{-2}}{6.93 	imes 10^{-3}} = \left(\frac{0.20}{0.10}ight)^a$

$2 = 2^a$

This implies that **$a = 1$**. The reaction is first-order with respect to $A$. Our master rate law simplifies beautifully to:

$r = k[A]^1$

### The Master Equation

With the rate law established, we can now unlock the rate constant, $k$. We can use the data from any of the experiments. Let's use Experiment I for simplicity.

$r_1 = k[A]_1$

$6.93 	imes 10^{-3} = k 	imes 0.10$

Solving for $k$:

$k = \frac{6.93 	imes 10^{-3}}{0.10} = 6.93 	imes 10^{-2} 	ext{ min}^{-1}$

Notice the units of $k$. Because it's a first-order reaction, the units are simply inverse time ($	ext{min}^{-1}$).

### The Final Countdown: Half-Life

The question asks for the time required to consume half of $A$. This is the very definition of **half-life** ($t_{1/2}$). For a first-order reaction, the half-life is a beautiful constant that does not depend on the initial concentration. It is given by the formula:

$t_{1/2} = \frac{\ln 2}{k} \approx \frac{0.693}{k}$

Substituting our calculated value of $k$:

$t_{1/2} = \frac{0.693}{6.93 	imes 10^{-2}}$

$t_{1/2} = 10 	ext{ min}$

It will take exactly 10 minutes for the concentration of $A$ to drop to half of its initial value. If you were to plot this, you would see a perfect exponential decay curve, dropping by half every 10 minutes, regardless of where you start on the curve. This elegant mathematical predictability is what makes chemical kinetics so powerful!

Select Answer:

Visualized Solution

\text{Objective: Half-life of } A

\text{Finding Order w.r.t } B

\text{Order of } B

\text{Finding Order w.r.t } A

\text{Order of } A

\text{Calculating Rate Constant } k

\text{Calculating Half-Life } t_{1/2}

\text{The Way Forward}

The Sigma Insight: Order and Molecularity

Decoding the Kinetic Data

The Power of Isolation

Unveiling the Order of A

The Master Equation

The Final Countdown

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